230 ANALYTICAL MECHANICS Therefore and v = ~ g sm a. o Thus both the linear and the angular accelerations of the cylinder are constant. Therefore the equations of the motion are the following: o - gt sin a, o s = vQt + - gtf2 sin a, o 2;2 = y2+-o -Tg co0 + ;j— <7 sin or, — o a 4 7T~ oa sn a 2. A wheel moves down an inclined groove with its axle rolling along the groove without slipping. Discuss the motion. Let a — the radius of the axle. b = the radius of the wheel. m'= the mass of that part of the axle which projects out from the wheel. m = the mass of the rest of the moving system. M = m + mf. Suppose the wheel to be a solid disk with a thickness equal to half the total length of the axle. Then if both the wheel and the axle are of the same FIG. 116. material the relation — = — m Therefore m' ; M and m = TORQUE METHOD.—Considering the moments about the element of contact we obtain the following for the torque equation: I(a = Mga sin a, (Ic+Ma?) o) — Mga sin a, (Ic + Maz) v = Mga2 sin a, where Ic denotes the moment of inertia of the moving system about its own axis. M. ENERGY METHOD. — Supposing the wheel to start from rest we obtain + $ Jcco2 = Mgs sin a, /c2 = Mussina, (Ic + Ma*) v2 = Mga*s sin a, (Ic + Ma2) vv = Mga2s sin a, (Ic + Ma2) v = M#a2 sin a.